SNVA994A February   2022  – March 2023 LM5157 , LM5157-Q1 , LM51571-Q1 , LM5158 , LM5158-Q1 , LM51581 , LM51581-Q1

 

  1.   1
  2.   Trademarks
  3. 1Introduction
  4. 2Example Application
  5. 3Calculations and Component Selection
    1. 3.1 Switching Frequency
    2. 3.2 Transformer Selection
      1. 3.2.1 Maximum Duty Cycle and Turns Ratio Selection
      2. 3.2.2 Primary Winding Inductance Selection
    3. 3.3 Slope Compensation Check
    4. 3.4 Diode Selection
    5. 3.5 Output Capacitor Selection
    6. 3.6 Input Capacitor Selection
    7. 3.7 UVLO Resistor Selection
    8. 3.8 Control Loop Compensation
      1. 3.8.1 Crossover Frequency (fcross) Selection
      2. 3.8.2 RCOMP Selection
      3. 3.8.3 CCOMP Selection
      4. 3.8.4 CHF Selection
  6. 4Component Selection Summary
    1. 4.1 Application Circuit
    2. 4.2 Bill of Materials
  7. 5Small Signal Frequency Analysis
    1. 5.1 Flyback Regulator Modulator Modeling
    2. 5.2 Compensation Modeling
  8. 6Revision History

Slope Compensation Check

According to the theory of the peak current mode control, the slope of the compensation ramp must be greater than half of the sensed inductor current falling slope to prevent subharmonic oscillation at high duty cycle. Therefore, the following inequality in Equation 9 should be satisfied.

Equation 9. 0.5×VLOAD+VF-VSUPPLYLM×ACS×Margin<500mV×fsw 

where

  • ACS is the equivalent current sensing gain.
  • 500mV is the slope compensation peak voltage.

Typically 82% of the sensed inductor current falling slope is an optimal value of the slope compensation, which reflects to a margin of 1.6. If the inequality is failed, the inductance value of LM must be increased so that the falling slope can be smaller. If the LM inductance value is changed the peak current must be re-calculated and the device selection must be re-examined. In this example, the inequality is verified in Equation 10, Equation 11, and Equation 12.

Equation 10. 0.5×VLOAD+VF-VSUPPLYLM×ACS×Margin=0.5×10V+0.5V-8V8μH×0.095×1.6=23.75×103
Equation 11. 500mV×fsw=500mV×250kHz=125×103
Equation 12. 23.75×103<125×103